P = b+a+c for b solve for b
1 answer:
Answer:
Step-by-step explanation:
Notice that there are 3 terms on the right hand side. You want to isolate b in order to solve for it.
Explanation:
I find it less confusing to place term containing the variable you want on the left hand side. (As long as it is positive).
a
+
b
+
c
=
P
To isolate b
subtract a
and c from both sides
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![\begin{array}{l} \textsf{On the problem, the amount of salt in the tank,} \\ A(t), \textsf{changes overtime is given by the differential} \\ \textsf{equation} \\ \\ \footnotesize A'(t) = \left(\dfrac{4\ \textsf{gal}}{1\ \textsf{min}}\right)\!\!\left(\dfrac{1\ \textsf{lb}}{1\ \textsf{gal}}\right) - \left(\dfrac{2\ \textsf{gal}}{1\ \textsf{min}}\right)\!\!\left(\dfrac{A(t)\ \textsf{lb}}{10 + (4 - 2)t\ \textsf{gal}}\right) \\ \\ \textsf{There's no salt in the tank (fresh water) at the} \\ \textsf{start, so }A(0) = 0. \textsf{ The amount of solution in the} \\ \textsf{tank is given by }10 + (4 -2)t, \textsf{so the tank will} \\ \textsf{overflow once this expression is equal to the total} \\ \textsf{volume or capacity of the tank.} \\ \\ 10 + (4 - 2)t = 50 \\ \\ \textsf{Solving for }t,\textsf{ we get} \\ \\ \implies \boxed{t = 20\textsf{ mins}} \\ \\ A'(t) = 4 - \dfrac{2A(t)}{10 + 2t} \\ \\ A'(t) = 4 - \dfrac{1}{5 + t} A(t) \\ \\ A'(t) + \dfrac{1}{5 + t} A(t) = 4 \\ \\ \textsf{This is a linear ODE with integrating factor} \\ \mu (t) = e^{\int \frac{1}{5 + t}\ dt} = e^{\ln |5 + t|} = 5 + t \\ \\ \textsf{Multiplying this to the ODE, we get} \\ \\ (5 + t)A'(t) + A(t) = 4(5 + t) \\ \\ [(5 + t)A(t)]' = 20 + 4t \\ \\ (5 + t)A(t) = 20t + 2t^2 + C \\ \\ \textsf{Since }A(0) = 0, \textsf{ we get } C = 0. \\ \\ A(t) = \dfrac{2t^2 + 20t}{t + 5} \\ \\ A(t) = 2t + 10 - \dfrac{50}{t + 5} \\ \\ \textsf{So the function that gives the amount of salt at} \\ \textsf{any given time }t,\textsf{ is given by} \\ \\ \implies A(t) = 2t + 10 - \dfrac{50}{t + 5} \\ \\ \textsf{The amount of salt in the tank at the moment} \\ \textsf{of overflow or at }t = 20\textsf{ mins is equal to} \\ \\ A(20) = 2(20) + 10 - \dfrac{50}{20 + 5} \\ \\ \implies \boxed{A = 48\ \textsf{gallons}} \end{array}](https://tex.z-dn.net/?f=%20%5Cbegin%7Barray%7D%7Bl%7D%20%5Ctextsf%7BOn%20the%20problem%2C%20the%20amount%20of%20salt%20in%20the%20tank%2C%7D%20%5C%5C%20A%28t%29%2C%20%5Ctextsf%7Bchanges%20overtime%20is%20given%20by%20the%20differential%7D%20%5C%5C%20%5Ctextsf%7Bequation%7D%20%20%5C%5C%20%5C%5C%20%5Cfootnotesize%20A%27%28t%29%20%3D%20%5Cleft%28%5Cdfrac%7B4%5C%20%5Ctextsf%7Bgal%7D%7D%7B1%5C%20%5Ctextsf%7Bmin%7D%7D%5Cright%29%5C%21%5C%21%5Cleft%28%5Cdfrac%7B1%5C%20%5Ctextsf%7Blb%7D%7D%7B1%5C%20%5Ctextsf%7Bgal%7D%7D%5Cright%29%20-%20%5Cleft%28%5Cdfrac%7B2%5C%20%5Ctextsf%7Bgal%7D%7D%7B1%5C%20%5Ctextsf%7Bmin%7D%7D%5Cright%29%5C%21%5C%21%5Cleft%28%5Cdfrac%7BA%28t%29%5C%20%5Ctextsf%7Blb%7D%7D%7B10%20%2B%20%284%20-%202%29t%5C%20%5Ctextsf%7Bgal%7D%7D%5Cright%29%20%5C%5C%20%5C%5C%20%5Ctextsf%7BThere%27s%20no%20salt%20in%20the%20tank%20%28fresh%20water%29%20at%20the%7D%20%5C%5C%20%5Ctextsf%7Bstart%2C%20so%20%7DA%280%29%20%3D%200.%20%5Ctextsf%7B%20The%20amount%20of%20solution%20in%20the%7D%20%5C%5C%20%5Ctextsf%7Btank%20is%20given%20by%20%7D10%20%2B%20%284%20-2%29t%2C%20%5Ctextsf%7Bso%20the%20tank%20will%7D%20%5C%5C%20%5Ctextsf%7Boverflow%20once%20this%20expression%20is%20equal%20to%20the%20total%7D%20%5C%5C%20%5Ctextsf%7Bvolume%20or%20capacity%20of%20the%20tank.%7D%20%5C%5C%20%5C%5C%2010%20%2B%20%284%20-%202%29t%20%3D%2050%20%5C%5C%20%5C%5C%20%5Ctextsf%7BSolving%20for%20%7Dt%2C%5Ctextsf%7B%20we%20get%7D%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cboxed%7Bt%20%3D%2020%5Ctextsf%7B%20mins%7D%7D%20%5C%5C%20%5C%5C%20A%27%28t%29%20%3D%204%20-%20%5Cdfrac%7B2A%28t%29%7D%7B10%20%2B%202t%7D%20%5C%5C%20%5C%5C%20A%27%28t%29%20%3D%204%20-%20%5Cdfrac%7B1%7D%7B5%20%2B%20t%7D%20A%28t%29%20%5C%5C%20%5C%5C%20A%27%28t%29%20%2B%20%5Cdfrac%7B1%7D%7B5%20%2B%20t%7D%20A%28t%29%20%3D%204%20%5C%5C%20%5C%5C%20%5Ctextsf%7BThis%20is%20a%20linear%20ODE%20with%20integrating%20factor%7D%20%5C%5C%20%5Cmu%20%28t%29%20%3D%20e%5E%7B%5Cint%20%5Cfrac%7B1%7D%7B5%20%2B%20t%7D%5C%20dt%7D%20%3D%20e%5E%7B%5Cln%20%7C5%20%2B%20t%7C%7D%20%3D%205%20%2B%20t%20%5C%5C%20%5C%5C%20%5Ctextsf%7BMultiplying%20this%20to%20the%20ODE%2C%20we%20get%7D%20%5C%5C%20%5C%5C%20%285%20%2B%20t%29A%27%28t%29%20%2B%20A%28t%29%20%3D%204%285%20%2B%20t%29%20%5C%5C%20%5C%5C%20%5B%285%20%2B%20t%29A%28t%29%5D%27%20%3D%2020%20%2B%204t%20%5C%5C%20%5C%5C%20%285%20%2B%20t%29A%28t%29%20%3D%2020t%20%2B%202t%5E2%20%2B%20C%20%5C%5C%20%5C%5C%20%5Ctextsf%7BSince%20%7DA%280%29%20%3D%200%2C%20%5Ctextsf%7B%20we%20get%20%7D%20C%20%3D%200.%20%5C%5C%20%5C%5C%20A%28t%29%20%3D%20%5Cdfrac%7B2t%5E2%20%2B%2020t%7D%7Bt%20%2B%205%7D%20%5C%5C%20%5C%5C%20A%28t%29%20%3D%202t%20%2B%2010%20-%20%5Cdfrac%7B50%7D%7Bt%20%2B%205%7D%20%5C%5C%20%5C%5C%20%5Ctextsf%7BSo%20the%20function%20that%20gives%20the%20amount%20of%20salt%20at%7D%20%5C%5C%20%5Ctextsf%7Bany%20given%20time%20%7Dt%2C%5Ctextsf%7B%20is%20given%20by%7D%20%5C%5C%20%5C%5C%20%5Cimplies%20A%28t%29%20%3D%202t%20%2B%2010%20-%20%5Cdfrac%7B50%7D%7Bt%20%2B%205%7D%20%5C%5C%20%5C%5C%20%5Ctextsf%7BThe%20amount%20of%20salt%20in%20the%20tank%20at%20the%20moment%7D%20%5C%5C%20%5Ctextsf%7Bof%20overflow%20or%20at%20%7Dt%20%3D%2020%5Ctextsf%7B%20mins%20is%20equal%20to%7D%20%5C%5C%20%5C%5C%20A%2820%29%20%3D%202%2820%29%20%2B%2010%20-%20%5Cdfrac%7B50%7D%7B20%20%2B%205%7D%20%5C%5C%20%5C%5C%20%5Cimplies%20%5Cboxed%7BA%20%3D%2048%5C%20%5Ctextsf%7Bgallons%7D%7D%20%5Cend%7Barray%7D%20)


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Answer: M= -3/4
Step-by-step explanation:
Answer:
Step-by-step explanation:
hope this helps
Answer: The first, The second, and the number line one love!
Step-by-step explanation:
X=4 makes it true
Thank you hope this helps